Question

In the figure below, the hanging object has a mass of m₁ = 0.410 kg; the sliding block has a mass of m₂ = 0.765 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R₁ = 0.020 0 m, and an outer radius of R₂ = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is ?_k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of v_i = 0.820 m/s toward the pulley when it passes a reference point on the table.

(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away.
m/s

(b) Find the angular speed of the pulley at the same moment.
rad/s

Answer 1

a) work done = Change in K.E                               ----------------------------1

Change in K.E of system ( block m1+ block m2+ pulley)

   = ( 1/2)* m1* ( V_f² - V_i² ) + ( 1/2)* m2*( V_f²- V_i² ) + ( 1/2)* I * ( w_f² - w_i² )

= ( 1/2) * ( V_f² - V_i²   ) ( m1+m2) + ( 1/2) * I * ( w_f² - w_i² )                                       ---------------------A

I ( moment of interia) =   ( 1/2) * M * ( R1² + R₂² )

V = wr,     w = V/r , so w_f = V_f / R2     ,       w_i = V_i / R2

by putting all the values in eq A,

change in K.E = ( 1/2)* ( V_f² - V_i² ) * ( m1+m2) + ( 1/2) * ( 1/2 M * ( R₁² + R₂² ) ) * ( V_f² - V_i² / R₂² )

net work done ( due to gravitational force and due to friction force on m2) :

= m1*g*d - u_k *m2*g*d                                         ( W= force*displacement)

from eq 1:

(1/2)* ( V_f² - V_i² ) ( m1+m2) + ( 1/4) * M * ( R₁² + R₂² ) * ( V_f² - V_i² / R₂² ) = m1*g*d - u_k * m2*g*d

bu putting all the values and sloving it

V_f = ( velocity after 0.700 m distance) = 1.665 m/s

b) angular speed , w_f = V_f / R2                                     ( V = wr)

                                = 1.6 / 0.030

                             = 55.5 rad/s